4
$\begingroup$

Let $p$ be a prime and let $F$ be a free group of rank $d\geq 1$. Kostrikin [1] proved that the $d$-generated Burnside group $B=B(d,p)=F/F^p$ of exponent $p$ has a maximal finite quotient $\overline{B}=\overline{B(d,p)}=B/N$ where $N$ is the intersection of all the subgroups of $B$ of finite index. I understand that Kostrikin proved that the nilpotency class $c(d,p)$ of $\overline{B(d,p)}$ has an upper bound.

Question 1: Are lower bounds known for $c(d,p)$?

Question 2: Is $c(d,p)\geq d$ for $d\leq 4$ and $p>d$?

Question 3: For which $n$ is the $n$th section of the lower exponent-$p$ central series for $\overline{B(d,p)}$ isomorphic to the $n$th Lie power $L^n(V)$ where $V=({\mathbb F}_p)^d$?

[1] Kostrikin, A. L: On Lie rings with Engel's condition. Dokl. Akad. Nauk SSSR 108, no. 4 (1956) 580-582.

$\endgroup$
0

1 Answer 1

3
$\begingroup$

For Questions 1, 2. We know $c(d,2)=1$ for all $d$ and $c(1,3)=1$, $c(2,3)=2$ and $c(d,3)=3$ for all $d>2$. So suppose if necessary $p\geq 5$.

Take $G(d,p)$ to be the unitiriangular matrices of size $(d+1)\times (d+1)$ over the field of size $p$, where $p>d$. Then $G(d,p)$ is nilpotent of class $d$ and the exponent is $p$. The group can be generated by $d$ elements.

$\endgroup$
2
  • $\begingroup$ Thanks Alireza. A lower bound can be established with a `witness': if $G$ is $d$-generated, has exponent $p$, and class $c$, then $c(d,p)\geq c$. For example $c(2,p)\geq 2$ if $p>2$; take $G$ to be extraspecial of exponent $p$. Your group $G(d,p)$ does not have exponent $p$ if $d>3$, and its class is $d-1$ not $d$. For Question 2 it suffices to find witnesses $G_{d,p}$ for $d=3,4$ and all $p\geq5$. $\endgroup$
    – Glasby
    Commented Jul 1, 2015 at 14:36
  • 1
    $\begingroup$ Sorry Alireza. I think you wanted $G(d,p)$ to be $(d+1)\times(d+1)$ upper triangular matrices over ${\mathbb F}_p$ (not $d\times d$). Then $G(d,p)$ is a witness: $d$-generated, has exponent $p$ if $p> d$, and class $d$. Thus $c(d,p)\geq d$ for $p>d$. This answers Questions 1 and 2. $\endgroup$
    – Glasby
    Commented Jul 1, 2015 at 15:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .